HW 6-solutions - Quest problems and solutions - Ohms Law, circuits, resistors and capacitors PDF

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Quest problems and solutions - Ohms Law, circuits, resistors and capacitors...

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bausinger (nlb934) – HW 6 – shih – (55050) This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001

10.0 points

Consider the circuit consisting of identical bulbs.

1

From the figure, with all the bulbs having the resistance R, IA = ID = IE =

V R

V 2R The rankings of the currents shown here are also the corresponding rankings of the brightness. IB = IC =

002 10.0 points You have a 6.28 W, 225 Ω resistor. What is the maximum current that should be allowed in it?

E D

Correct answer: 0.167066 A. B

E

C A

Rank the brightness of the bulbs. You may find it helpful to work out the currents through the bulbs for some fixed value of R and V , then compare the currents. 1. A = B = C > D > E 2. A = B = C > D = E 3. C > B > A > D > E 4. A = C > B > D = E 5. E = D > A > B = C 6. A = D = E > B = C correct 7. B = C > A = D = E 8. A = D = E > B > C 9. A = B > C > D = E 10. A = B = C = D = E Explanation:

Explanation: Let : P = 6.28 W R = 225 Ω .

and

Power is P = I2 R P I2 = R r r 6.28 W P = I= = 0.167066 A . R 225 Ω 003 10.0 points Dielectric materials used in the manufacture of capacitors are characterized by conductivities that are small but not zero. Therefore, a charged capacitor slowly loses its charge by “leaking” across the dielectric. If a certain 4.32 µF capacitor leaks charge such that the potential difference decreases to half its initial value in 4.7 s, what is the equivalent resistance of the dielectric? Correct answer: 1.5696 × 106 Ω.

Explanation:

Let : t = 4.7 s and C = 4.32 µF = 4.32 × 10−6 F . By using q = C V , we immediately find out that when the potential across the capacitor

bausinger (nlb934) – HW 6 – shih – (55050) is halved, the charge on the capacitor is also halved; i.e., qf 1 = . qi 2 Also note that the charge drop has time dependence as q = q0 e−t/RC .

R= C ln

t 

R1 + R2 C 2 1 7. τ = (R1 + R2) C

6. τ =

8. τ = R2 C 9. τ =

Solving for R yields

10. τ = q0 q

.

1 R2 C 2 (R1 + R2) C

Explanation: In charging an R C circuit, the characteristic time constant is given by τ = R C , where R is the equivalent resistance:

Thus we have, for q0 = qi and q = qf , t 

 qi C ln qf (4.7 s) = (4.32 × 10−6 F) ln (2)

R=

R = R1 + R2 .

005 (part 2 of 2) 10.0 points S has been left at position a for a long time. It is then switched from a to b at t = 0. 3.8 MΩ

= 1.5696 × 106 Ω .

004 (part 1 of 2) 10.0 points

0.5 µF

1 MΩ

The switch S has been in position b for a long period of time. R3 C

R2 R1

S b E

a

When the switch is moved to position a, find the characteristic time constant. 1. τ =

1

R1 C p 2. τ = R1 R2 C 3. τ = R1 C 4. τ = √

1 R1 R2 C

5. τ = (R1 + R2) C correct

2

S b 0.5 V

1.5 MΩ

a

Determine the energy dissipated through the resistor R2 from t = 0 to t = ∞. Correct answer: 0.0130208 µJ. Explanation:

Let : E = 0.5 V , R1 = 1.5 MΩ , R2 = 1 MΩ , R3 = 3.8 MΩ , C = 0.5 µF . The total energy dissipated is R 2 +R 3 = Udissip

1 C E2 . 2

and

bausinger (nlb934) – HW 6 – shih – (55050) Since R2 and R3 are in series, the energy R2 dissipated by R2 is only a fraction of R2 + R3 the total energy: R2 Udissip

=



R2 R2 + R3



1 C E2 2



U

=

Z

Z0

D

x

.

B

0.9 m

1.

P BC

I(t)2 R2 dt

BB

∞

BA

BD

BD

2.

BB

BC P

BA

BA

0

1 R2 = C E2 R2 + R3 2 1 MΩ 1 = (0.5 µF) (0.5 V)2 1 MΩ + 3.8 MΩ 2

3.

BC

P BD

BB

= 0.0130208 µJ . BB 006 10.0 points Four long, parallel conductors carry equal 6 A currents, oriented at the corners of a square with sides of length of 0.9 m . A crosssectional view of the conductors is shown. The current direction is out of the page at points indicated by the dots and into the page at points indicated by the crosses.

×

C

Which diagram correctly denotes the directions of the components of the magnetic field from each conductor at the point P ?

∞

I 02 R2 e−2 t/[C (R2 +R3 )] dt 0   2  C (R2 + R3 ) E = R2 − 2 R2 + R3 ∞  × e−2 t/[C (R2 +R3 )]  =

y ×

P

(Power consumption consideration provides an independent check on the fraction used.) Since the two resistors are in series they share a common current I. The corresponding power consumptions by R2 and R3 are, respectively, P2 = I 2 R2 and P3 = I 2 R3. This shows the correctness of the fraction R2 P2 = . R2 + R3 P2 + P3 Alternate solution: More formally, noting that the initial current E is I0 = , the total energy dissipated R2 + R3 by R2 is R2

A

3

4.

BA P BD

BC

bausinger (nlb934) – HW 6 – shih – (55050)

BB

BD

coil that will convert electric energy to heat at a rate of 229 W for a current of 2.1 A. Determine the resistance of the coil.

BA

5.

P

Correct answer: 51.9274 Ω.

BC

Explanation: Let : P = 229 W I = 2.1 A.

BB

BC

6.

Power is

BA

so the resistance of the coil is

BD

R= Explanation: Applying the right-hand rule, the thumb of your right hand points in the direction of the current and your fingers curl around the wire in the direction of the magnetic field’s circular path: D

×

229 W P = 51.9274 Ω . = 2 (2.1 A)2 I

008 (part 2 of 2) 10.0 points The resistivity of the coil wire is 1.42 × 10−6 Ω · m, and its diameter is 0.161 mm. Find its length. Correct answer: 0.744475 m. Explanation:

BA

BD Let : ρ = 1.42 × 10−6 Ω · m

P BB

and

r = 0.0805 mm = 8.05 × 10−5 m .

BC ×

B

C

We use the formula for resistance as a function of resistivity ρ, length ℓ and crosssectional area A: ℓ ℓ =ρ π r2 A (51.9274 Ω) π (8.05 × 10−5 m)2 R ℓ = π r2 = ρ 1.42 × 10−6 Ω · m

R=ρ

Applying superposition, A

and

P = I2 R ,

correct

P

A

4

D

×

= 0.744475 m .

BB

BC P BA

B

BD ×

C

007 (part 1 of 2) 10.0 points Suppose that you want to install a heating

009 10.0 points A 37 m length of coaxial cable has a solid cylindrical wire inner conductor with a diameter of 3.124 mm and carries a charge of 6.71 µC. The surrounding conductor is a cylindrical shell and has an inner diameter of 9.02 mm and a charge of −6.71 µC. What is the capacitance of this cable? Assume the region between the conductors is

bausinger (nlb934) – HW 6 – shih – (55050) air. The value of the Coulomb constant is 8.98755 × 109 N · m2 /C2. Correct answer: 1.94129 nF.

Correct answer: 2.01024 µF. Explanation:

Explanation: Let :

Z

b

~ V =− E · d~s = −2 ke λ a   b Q . = −2 ke ln ℓ a

Z

a

b

a

dr r

EB

1   b ln a 37 m = 2 (8.98755 × 109 N · m2/C2 )   1 × 109 nF 1  · ×  9.02 mm 1F ln 3.124 mm Q ℓ = V 2 ke

= 1.94129 nF . 010 (part 1 of 2) 10.0 points Consider the group of capacitors shown in the figure. 1.64 µF 5.69 µF a b

3.46 µF

7.96 µF c d

15.3 V

C1 b

C3

c

C4 d

C2 and C3 are parallel, so C23 = C2 + C3 = 1.64 µF + 3.46 µF = 5.1 µF . C1 , C23, and C4 are in series, so   1 1 1 −1 Cda = + + C1 C23 C4 −1  1 1 1 + + = 5.69 µF 5.1 µF 7.96 µF = 2.01024 µF . 011 (part 2 of 2) 10.0 points Find the charge on 1.64 µF capacitor on the top. Correct answer: 9.89037 µC. Explanation: Capacitors in series have the same charge: Q1 = Q4 = Cda Vtotal = (2.01024 µF)(15.3 V) = 30.7566 µC . The voltage drops across C1 and C4 are Q1 30.7566 µC = = 5.40539 V C1 5.69 µF Q4 30.7566 µC = 3.8639 V . V4 = = C4 7.96 µF V1 =

Find the equivalent capacitance between points a and d.

and

C2

Q . ℓ

The capacitance of a cylindrical capacitor is given by C≡

Let : C1 = 5.69 µF , C2 = 1.64 µF , C3 = 3.46 µF , C4 = 7.96 µF , EB = 15.3 V .

ke = 8.98755 × 109 N · m2/C2 , Q = 6.71 µC , ℓ = 37 m , a = 3.124 mm , and b = 9.02 mm .

The charge per unit length is λ ≡

5

and

bausinger (nlb934) – HW 6 – shih – (55050)

6

The remaining voltage is

r

R ℓ

Vremain = Vtotal − V1 − V4 = 15.3 V − 5.40539 V − 3.8639 V = 6.03071 V , and is the same across the parallel capacitors: Q2 = C2 Vremain = (1.64 µF)(6.03071 V) = 9.89037 µC .

012 10.0 points Consider a long, uniformly charged, cylindrical insulator of radius R with charge density 1.5 µC/m3.

The flux leaving the ends of the Gaussian cylinder is negligible, and the only contribution to the flux is from the side of the cylinder. Since the field is perpendicular to this surface, the flux is Φs = 2 π r ℓ E , and the charge enclosed by the surface is Qencl = π r2 ℓ ρ .

R 3 cm

What is the magnitude of the electric field inside the insulator at a distance 3 cm < R from the axis? The permittivity of free space is 8.8542 × 10−12 C2 /N · m2 and the volume of a cylinder with radius r and length ℓ is V = π r2 ℓ . Correct answer: 2541.17 N/C. Explanation:

Let : r = 3 cm = 0.03 m , ρ = 1.5 µC/m3 , = 1.5 × 10−6 C/m3 , and ǫ0 = 8.8542 × 10−12 C2 /N · m2 . Consider a cylindrical Gaussian surface of radius r and length ℓ much less than the length of the insulator so that the component of the electric field parallel to the axis is negligible.

Using Gauss’ law, Qenc Φs = ǫ0 π r2 ℓ ρ 2πrℓE = ǫ0 ρ E= r 2 ǫ0   1.5 × 10−6 C/m3 (0.03 m) = 2 (8.8542 × 10−12 C2 /N · m2 ) = 2541.17 N/C .

013 10.0 points A capacitor is constructed from two metal plates. The bottom portion is filled with air and the top portion is filled with material of dielectric constant κ. The plate area in the top region is the same as that in the bottom region. Neglect edge effects. κ

E

bausinger (nlb934) – HW 6 – shih – (55050) Ut of energy stored in Ub the top portion to the bottom portion. Determine the ratio

Ut 1 1. = 2 κ Ub Ut = 2κ + 1 2. Ub Ut 1 3. = κ+1 Ub Ut 1 4. = Ub 2κ Ut = 2κ 5. Ub Ut = κ correct 6. Ub Ut 1 7. = 2κ + 1 Ub Ut 8. = κ2 Ub Ut 1 9. = Ub κ Ut = κ+1 10. Ub Explanation: For a capacitor with dielectric κ, κ ǫ0 A d and the energy stored in a charged capacitor is 1 1 Q2 1 . U = QV = V 2C = 2 2 2 C The capacitance is determined solely by the dielectric constant of the material and geometric configuration of the capacitor. The bottom portion and the top portion may be treated as two capacitors connected in parallel: C=

Ct

Vb = Vt = V , so 1 2 V Ct Ut = 2 1 2 Ub V Cb 2 014

=

Ct = κ. Cb

10.0 points

A positively charged particle moving at 45◦ angles to both the y-axis and z-axis enters a magnetic field (pointing out of of the page), as shown. ˆı is in the x-direction, ˆ is in the y-direction, and kˆ is in the z-direction. +q ~ B v

x ~ B

×

y

z What is the initial direction of deflection?  1  ˆ 1. Fb = √ −k − ˆı 2 1 2. Fb = √ (+ˆı + ˆ) 2  1  ˆ 3. Fb = √ +k − ˆ 2   1 4. Fb = √ +kˆ + ˆı 2 1 5. Fb = √ (−ˆı − ˆ) 2   1 −kˆ − ˆ correct 6. Fb = √ 2   1 7. Fb = √ +kˆ + ˆ 2 ~ 8. F = 0 ; no deflection 1 9. Fb = √

Cb

7

1 10. Fb = √

2



2

(+ˆı − ˆ)

Explanation:

 +kˆ − ˆı

bausinger (nlb934) – HW 6 – shih – (55050) ~ , B ~ = B (−ˆ) , The force is ~F = q ~v × B  1  ~v = √ v −ˆ + ˆk , and q > 0 , so 2 ~ = +|q| ~v × B ~ F h  i 1 = +|q| √ v B −ˆ + kˆ × (−ˆı) 2   1 = +|q| √ v B −kˆ − ˆ 2   b = √1 −kˆ − ˆ . F 2

8...